The Kadison-Singer and Paulsen Problems in Finite Frame Theory

18 Peter G. Casazzalooking question turns out to be much deeper than it looks, and as we willnow see, it is equivalent to PC.Conjecture 11. There exists an ɛ > 0 so that for large K, for all N and allequal norm Parseval frames {ϕ i } KNi=1 for lN 2 , there is a J ⊂ {1, 2, . . . , KN} sothat both {ϕ i } i∈J and {ϕ i } i∈J c have lower frame bounds which are greaterthan ɛ.The ideal situation would be for Conjecture 11 to hold for all K ≥ 2. Inorder for {ϕ i } i∈J and {ϕ i } i∈J c to both be frames for l N 2 , they at least haveto span l N 2 . So the first question is whether we can partition our frame intospanning sets. This follows from a generalization of the Rado-Horn theorem.See the chapter on independence and spanning properties of frames.Proposition 4. Every equal norm Parseval frame {ϕ i } KN+Li=1 , 0 ≤ L < Nfor l N 2 can be partitioned into K linearly independent spanning sets plus alinearly independent set of L elements.The natural question is whether we can do such a partition so that eachof the subsets has good frame bounds, i.e., a universal lower frame bound forall subsets. Before addressing this question, we state another conjecture.Conjecture 12. There exists ɛ > 0 and a natural number r so that for allN, all large K and all equal norm Parseval frames {ϕ i } KNi=1 in lN 2 there is apartition {A j } r j=1 of {1, 2, . . . , KN} so that for all j = 1, 2, . . . , r the Besselbound of {ϕ i } i∈Aj is ≤ 1 − ɛ.We will now establish a relationship between our conjectures and PC.Theorem 11. (1) Conjecture 11 implies Conjecture 12.(2) Conjecture 12 is equivalent to PC.Proof. (1): Fix ɛ > 0, r, K as in Conjecture 11. Let {ϕ i } KNi=1 be an equal normParseval frame for an N-dimensional Hilbert space H N . By Naimark’s Theoremwe may assume there is an orthogonal projection P on l KN2 with P e i = ϕ ifor all i = 1, 2, . . . , KN. By Conjecture 11, there is a J ⊂ {1, 2, . . . , KN} sothat {P e i } i∈J and {P e i } i∈J c both have a lower frame bound of ɛ > 0. Hence,for x ∈ H M = P (l KN2 ),∑KN‖x‖ 2 = |〈x, P e i 〉| 2 = ∑ |〈x, P e i 〉| 2 + ∑ |〈x, P e i 〉| 2i=1i∈Ji∈J c≥ ∑ i∈J|〈x, P e i 〉| 2 + ɛ‖x‖ 2 .That is, ∑ i∈J |〈x, P e i〉| 2 ≤ (1−ɛ)‖x‖ 2 . So the upper frame bound of {P e i } i∈J(which is the norm of the analysis operator (P Q J ) ∗ for this frame) is ≤ 1 − ɛ.Since P Q J is the synthesis operator for this frame, we have that ‖Q J P Q J ‖ =